1. Hmmmm. Cannot be determined. Assuming one is acid and one is base, the immediate question is that a mole does not neutralize a mole: one could be di or tri protic. You have to have a balanced equation. Frankly, I like to work with Normal solution on these.
2. Look at your solubility rules, for instance: on the first, all potassium compounds are soluble, but not all mercury: is Mercury(II)sulfate soluble?
Memorize those rules, millions before you have done the same.
Could someone please help me with these two questions. I am having trouble.
1. What is the molarity of a solution if 48.0 mL of a 0.220 M solution is required to neutralize a 25.0-mL sample of the solution?
2. The mixing of which pair of reactants will result in a precipitation reaction?
K 2SO4(aq) + Hg2 (NO3)2(aq)
CsI(aq) + NaOH(aq)
HCl(aq) + Ca(OH)2(aq)
NaNO3(aq) + NH4Cl(aq)
3 answers
Ok, great! Thank you.
Sorry about question 1, I left out the full question. This is it:
What is the molarity of a NaOH solution if 48.0 mL of a 0.220 M H2SO4 solution is required to neutralize a 25.0-mL sample of NaOH solution?
Sorry about question 1, I left out the full question. This is it:
What is the molarity of a NaOH solution if 48.0 mL of a 0.220 M H2SO4 solution is required to neutralize a 25.0-mL sample of NaOH solution?
2NaOH + H2SO4 --> Na2SO4 + H2O
moles H2SO4 = M x L
moles NaOH = 2x moles H2SO4
Molarity NaOH = moles NaOH/L NaOH.
moles H2SO4 = M x L
moles NaOH = 2x moles H2SO4
Molarity NaOH = moles NaOH/L NaOH.
1 answer